The big idea: A planet stays in orbit because the Sun's gravity constantly pulls it inward.
That inward pull bends the planet's path into a closed orbit instead of letting it fly off in a straight line.
Kepler's three laws describe the shape, speed and timing of these orbits.
[Diagram: phys-field-lines] - Available in full study mode
Kepler's three laws in plain words: 1st law: orbits are ellipses (slightly squashed circles), with the Sun at one focus.
2nd law: a planet moves faster when it is closer to the Sun and slower when it is farther away.
3rd law: the period T and the orbit radius r are linked — bigger orbits take longer.
Kepler's third law links how long an orbit takes to how big it is. The period squared is proportional to the orbit radius cubed:
- orbital period — time for one full orbit (s)
- orbital radius — distance from the central body (m)
- universal gravitational constant (6.67 × 10⁻¹¹ N m² kg⁻²)
- mass of the central body being orbited (kg)
Where it comes from (both pieces ARE given): Gravity provides the centripetal pull, so the given field equation g = GM ÷ r² equals the given centripetal acceleration a = 4π²r ÷ T².
Setting them equal and rearranging gives T² = 4π²r³ ÷ (GM).
You rarely need the constants: the useful idea is just T² is proportional to r³.
- gravitational field strength (N kg⁻¹)
- gravitational force on the orbiting mass (N)
- mass of the orbiting body (kg)
- mass of the central body (kg)
- distance from the centre of the central body (m)
Comparing two orbits — the shortcut: Because T² ÷ r³ is the same for every body orbiting the same central mass, two orbits A and B obey:
T_{A}² ÷ r_{A}³ = T_{B}² ÷ r_{B}³
The constant 4π² ÷ (GM) cancels, so you never need G or M — just the ratios.
Worked example — period of a larger orbit
Planet A orbits a star with period 2.0 years at radius r. Planet B orbits the same star at radius 4r. Find the orbital period of planet B.
Solution
- Start with Kepler's third law as a ratio (T² ÷ r³ is the same for both):
- Rearrange to make TB the subject:
- Put in the numbers (TA = 2.0, rB/rA = 4):
- Take the square root — keep the unit:
Final answer
TB = 16 years — a 4× bigger orbit takes 8× longer (since 4³ = 64, and √64 = 8).
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How this is tested: Kepler's laws are the tool the orbits questions are built on.
- Paper 1A: a quick MCQ — recognise that T² ∝ r³, or pick the right exponents in T_{ⁿ} ∝ r_{ᵐ}. - Paper 2: calculate a ratio of two periods from their radii (or two radii from their periods) using TA² ÷ rA³ = TB² ÷ rB³, or state Kepler's first law, or explain the speed change from the second law.
Classic trap: forgetting the powers — it is T squared and r cubed, not T and r.
Ratio method (no G or M needed): For two bodies round the same central mass:
T_{A}² ÷ r_{A}³ = T_{B}² ÷ r_{B}³
Rearrange for whichever quantity is missing. The constant cancels, so you never need the star's mass.
IB-style question — (a) ratio of orbital radii
Two planets, X and Y, orbit the same star. Planet X has an orbital period of 1.0 year; planet Y has a period of 8.0 years. Find the ratio of their orbital radii, rY : rX.
Solution
- Both orbit the same star, so Kepler's third law as a ratio applies:
- Rearrange for the ratio of the radii cubed:
- Put in the numbers (TY/TX = 8.0/1.0 = 8):
- Take the cube root:
Final answer
rY : rX = 4.0 : 1 — planet Y orbits four times farther out.
IB-style question — (b) state Kepler's first law
State Kepler's first law of planetary motion.
Solution
- The first law is about the shape of an orbit, not its timing.
- Each planet moves in an ellipse with the Sun at one focus of that ellipse.
Final answer
Kepler's first law: each planet orbits the Sun in an ellipse, with the Sun at one focus of the ellipse.